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mamaluj [8]
3 years ago
15

I NEED HELP WITH A PROBLEM ​

Mathematics
1 answer:
postnew [5]3 years ago
6 0

Answer:

the first one I guess

Step-by-step explanation:

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For a quadratic function, which characteristics of its graph is equivalent to the zero of the function? a) y-intercept b) maximu
Sonja [21]

Answer:

d) x-intercept

Step-by-step explanation:

The zeros of a function are the values of x that make y = 0.

Thus, they are the points where the graph crosses the x-axis, that is, the x-intercepts.

a) is wrong. the y-intercept is simply the value of y when x = 0.

b) and c) are wrong. A quadratic can have only one maximum or one minimum; it can't have both. And they are not zeros except in the special case of y = ±ax².

The diagram below is the graph of y = x² + 3x - 46. It shows the zeros at x = -23 and x = 20. The minimum and the y-intercept are not zeros of the function.

3 0
3 years ago
The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units
ad-work [718]
Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&A&(~ 2 &,& 8~) 
%  (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&A&(~ 2 &,& 8~) 
%  (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}

so, add AC + AB + CB, and that's the perimeter of the triangle.

8 0
3 years ago
I need help quick!!<br><br> you can have brainliest this is algerbra btw
padilas [110]

Answer:

Step-by-step explanation:

Day 1 = 2 birds

day 2 = 3 birds

day 3 = 9 birds

day 4 = 8 birds

day 5 = 1 bird

On day 1 the points are (1,2)

On day 5 the points are (5,1)

(I am picking these for the points because I don't know your options for the selecting area.)

we use this formula y2-y1/x2-x1

2-5/1-1

3/0

our answer is infinite which is the relation and this is an infinite function

5 0
2 years ago
Read 2 more answers
Point M is the midpoint of AB . Through point M a line segment XY is drawn, so that both XA and YB are perpendicular to AB . Pro
GaryK [48]

Answer:

See the proof below

Step-by-step explanation:

Let the line AB be a straight line on the parallelogram.

A dissection of the line (using the perpendicular line X) gives:

AY ≅ BX

Another way will be using the angles.

The angles are equal - vertically opposite angles

Hence the line  AY ≅ BX (Proved)

8 0
3 years ago
Read 2 more answers
Sketch the region that is common to the graphs of x ≥ 2,
Oliga [24]

a. Find the graph of their common region in the attachment

b. The area of the common region of the graphs is 8 units²

<h3>a. How to sketch the region common to the graphs?</h3>

Since we have x ≥ 2, y ≥ 0, and x + y ≤ 6, we plot each graph separately and find their region of intersection.

  • The graph of x ≥ 2 is the region right of the line x = 2.
  • The graph of y ≥ 0 is the region above the line y = 0 or x-axis.
  • To plot the graph of x + y ≤ 6, we first plot the graph of x + y = 6 ⇒ y = -x + 6. Then the graph of x + y ≤ 6 is the graph of y ≤ - x + 6.

So, the graph of x + y ≤ 6 is the region below the line y = - x + 6

From the graph, the regions intersect at (2, 0), (2, 4) and (6, 0)

Find the graph of their common region in the attachment

<h3>b. The area of the common region</h3>

From the graph, we see that the common region is a right angled triangle with

  • height = 4 units and
  • base = 4 units

So, its area = 1/2 × height × base

= 1/2 × 4 units × 4 units

= 1/2 × 16 units²

= 8 units²

So, the area of the common region of the graphs is 8 units²

Learn more about region common to graphs here:

brainly.com/question/27932405

#SPJ1

6 0
2 years ago
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